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Question Number 49559 by Rio Michael last updated on 07/Dec/18

Commented by Rio Michael last updated on 07/Dec/18

$T h e f i g u r e b e l o w s h o w s a n o b j e c t o f m a s s 60 k g p u l l e d u p$ $b y a n e f f o r t o f 4000 J t h r o u g h a n i n c l i n e d p l a n e 30 ° t o t h e$ $h o r i z o n t a l .$ $a) S k e t c h t h e d i a g r a m s h o w i n g a l l f o r c e s a c t i n g o n t h e o b j e c t$ $a s i t m o v e s u p t h e i n c l i n e d p l a n e .$ $b) T h e o b j e c t i s a l l o w e d t o f a l l t h r o u g h a v e r t i c a l h e i g h t o f 6 m$ $c a l c u l a t e t h e c h a n g e i n p o t e n t i a l e n e r g y .$ $c) C a l c u l a t e t h e e f f i c i e n c y o f t h e s y s t e m a b o v e h e n c e s t a t e$ $w h y i t s l e s s t h e n 100 % .$
	Answered by afachri last updated on 07/Dec/18

	Commented by afachri last updated on 08/Dec/18

	$sir, pardon me . i did a mistake in the$ $diagram . the arrow of w c o s 30$ $should be w, and w arrow should be$ $w c o s 30 . i am so sorry .$
	Commented by Rio Michael last updated on 08/Dec/18

	$n o p r o b l e m s i r$
	Commented by Rio Michael last updated on 08/Dec/18

	$t h a n k s s i r i s a w t h a t a l r e a d y$
	Commented by afachri last updated on 07/Dec/18

	$your welcome Sir . the other$ $answer below .$
	Answered by afachri last updated on 07/Dec/18

	$B) {let}^{'} s say the gravity conts : 10 m / s^{2}$ $△ P o t e n t i a l e n e r g y = m . g . △ h$ $= (60 Kg) (10 m / s^{2}) (6 m)$ $= 3600 J$ $C) find first : w s i n (30) and F$ $w s i n (30 °) = m \times g \times s i n (30 °)$ $= 60 \times 10 \frac{1}{2}$ $= 300 J$ $and,$ $F = \frac{W}{s}$ $= (4000 J) / 12 m = \frac{1000}{3} N$ $so, the efficiency of system will be :$ $e = \frac{w s i n (30 °)}{F} \times 100 %$ $= 90 % \Rightarrow reason : cause applied effort (4000 J; 300 N)$ $is over than the force needed (\frac{1000}{3} N)$
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